Zaccagliaodi

2022-02-26

Evaluate:
$\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}+1}{n\cdot {2}^{n-1}}$

Capodarcod0f

What you can do is split it in two
$\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}+1}{n\cdot {2}^{n-1}}=\sum _{n=1}^{+\mathrm{\infty }}n{\left(\frac{1}{2}\right)}^{n-1}+\sum _{n=1}^{+\mathrm{\infty }}\frac{{\left(\frac{1}{2}\right)}^{n-1}}{n}$
$\frac{1}{n{2}^{n-1}}=2\frac{1}{n{2}^{n}}=2{\int }_{0}^{\frac{1}{2}}{x}^{k-1}dx$
Hence
$\sum _{k=1}^{N}\frac{1}{{2}^{k-1}}=2\sum _{k=1}^{N}{\int }_{0}^{\frac{1}{2}}{x}^{k-1}dx$
$=2{\int }_{0}^{\frac{1}{2}}\sum _{k=1}^{N}{x}^{k-1}dx$
$=2{\int }_{0}^{\frac{1}{2}}\frac{1-{x}^{N}}{1-x}dx$
From this, you have
$\sum _{k=1}^{N}\frac{1}{k{2}^{k-1}}=2{\int }_{0}^{\frac{1}{2}}\frac{dx}{1-x}-2{\int }_{0}^{\frac{1}{2}}\frac{{x}^{N}}{1-x}dx$
Making $N\to +\mathrm{\infty }$, I let you show that
${\int }_{0}^{\frac{1}{2}}\frac{{x}^{N}}{1-x}dx\to 0$
Hence
$\sum _{k=1}^{+\mathrm{\infty }}\frac{1}{k{2}^{k-1}}=2{\int }_{0}^{\frac{1}{2}}\frac{dx}{1-x}=\mathrm{log}\left(4\right)$

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